A ballon whose volume is `5.0 L` contains `7 g of N_(2)`. What mass of the `H_(2)` gas must be added to the ballon to expand its volume to `10 L` at the same temperature and pressure ?
Strategy : We know the two volumes at the same temperatures and pressure. We can find the moles of `N_(2)` through its mass . To calculate the total moles of `N_(2)` and `H_(2)` , we apply Avogadro's law. Finally , we find the mass of `H_(2)` through its moles.

Explanation

List the data :
`V_(1) = 5.0 L , V_(2) = 10 L`
`n_(1) = n_(N_(2)) = ("Mass"_(N_(2)))/("Molar mass"_(N_(2)))`
`n_(2) = n_(N_(2)) + n_(H_(2))`
`= (7.0 g N_(2))/(28.0 g N_(2) "mol"^(-1))`
` = 0.25 "mol" N_(2)`
Using Eq.(5.17), we obtain
`(V_(1))/(n_(1)) = (V_(2))/(n_(2))`
`(5.0 L)/(0.25 mol) = (10 L)/(n_(2))`
`:. n_(2) = (0.25 "mol") ((10 L)/(5.0 L)) = 0.5 "mo"l`
This is the total amount of gas required for the volume to be `10 L`. Thus ,
`n_(H_(1)) = n_(2) - n_(1) = (0.5 "mol") - (0.25 "mol")`
`= 0.25 "mol"`
Hence , the mass of `H_(2)` that must be added ,
`m_(H_(1)) = (n_(H_(1))) ("Molar mass"_(H_(1)))`
`= (0.25 "mol") (2.0 g "mol"^(-1)) = 0.5 g`